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B Definition of the limit of a sequence

  1. Jul 27, 2016 #1

    Math_QED

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    The definition of a limit of a sequence,
    if the limit is finite, is:

    lim n >infinity un (un is a sequence) = l

    <=>

    ∀ε> 0, ∃N: n > N => |un - l| < ε

    This just means that un for n > N has to be a number for which: l -ε < un < l + ε

    Now, I'm wondering, can't we just say:

    n > N => |un -l| < kε with k a real number larger than zero?

    Thanks in advance
     
  2. jcsd
  3. Jul 27, 2016 #2

    fresh_42

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    What for? There is no mathematical difference between ##ε## and ##kε## for a constant ##k##. Why do you want to complicate something without need that often is hard to understand to newbies anyway.
     
  4. Jul 28, 2016 #3

    micromass

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    I disagree that it is without need. You often see in mathematical derivations that something is ##<3\varepsilon## and that that's good enough. Such proofs are even known as ##3\varepsilon##-proofs. I do agree that what's in the OP shouldn't be the standard definition as it already is hard enough. But it should be made much more clearer in analysis books that ##<k\varepsilon## is good enough.
     
  5. Jul 28, 2016 #4

    PeroK

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    When people are learning limits often a lot of complexity comes about for the case where ##\epsilon## is not small. Also, sometimes it complicates things to get the ##< \epsilon## in the definition and it's simpler to get ##\le \epsilon##. I always thought a better (equivalent) definition would be.

    ##\exists \epsilon_0 > 0: \forall \epsilon \in (0, \epsilon_0), \exists N: n > N \Rightarrow |u_n - L| \le \epsilon##

    In fact, as I don't have to sit exams, this is the one I use!

    You could, if you wanted to, add your ##k##:

    ##\exists k, \ \epsilon_0 > 0: \forall \epsilon \in (0, \epsilon_0), \exists N: n > N \Rightarrow |u_n - L| \le k\epsilon##
     
  6. Jul 28, 2016 #5

    fresh_42

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    I learned the notation ##\forall \epsilon \exists N(\epsilon)## or ##\forall \epsilon \exists N_\epsilon## so there has never been a doubt that a the find of ##N## depends on the choice of ##\epsilon## and a constant factor can easily be transformed into ##N_\epsilon##.

    I find an additional "variable" in the definition would only cause more difficulties, such as to think the small sized ##\epsilon## could be blown up beyond any natural number. Although still correct, it's not helpful. (IMO, but I don't want hell to break lose ...:wink:)
     
  7. Jul 28, 2016 #6

    fresh_42

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    I agree, however, wouldn't that simply transform the difficulties to later concepts, because it then makes the difference between an open and a closed neighborhood. I always thought this to be the reason for the strict < . Otherwise one has to explain why all of a sudden we change from closed to open.
     
  8. Jul 28, 2016 #7

    micromass

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    You're right. But when I first learned analysis at university, one of the very first exercises we did was to show that we could replace all notions of ##<## with ##\leq##. I remember that I found it somewhat difficult to grasp, and I didn't really see its importance. In retrospect, I'm happy that we did this.
     
  9. Jul 28, 2016 #8

    Math_QED

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    Yes, this is exactly why I made this thread. In the following proof, we encounter something similar:

    Theorem: If limn-> ∞un = l1 and limn-> ∞vn = l2

    limn-> ∞(un + vn) = l1 + l2

    Proof: We can say for all ε>0

    ∃N: n> N => |un-l1|<ε
    ∃N': n> N' => |vn-l2|<ε

    Now, we observe, that:

    |un + vn - (l1 + l2)| = |(un - l1) + (vn - l2)| =< |un - l1| + |vn - l2|

    So, for n > max{N, N'}, we find:

    |un + vn - (l1 + l2)| = |(un - l1) + (vn - l2)| =< |un - l1| + |vn - l2| < ε + ε

    => |un + vn - (l1 + l2)| < 2ε

    Now, the book does something I'm unsure about:

    They say, choose ε' > 0, and let ε = ε'/2. Then we have, for all ε' > 0

    n > max {N, N'} => |un + vn - (l1 + l2)| < ε' and what we wanted to proof follows.

    It makes sense to me that |un + vn - (l1 + l2)| < 2ε would be enough to proof, as Micromass already said (that's why I created this thread), but I do not fully understand why we can say that ε = ε'/2? Is this, because ε>0 => ε'/2 > 0?

    Probably this is a dumb question, but I'm self studying this to prepare for university mathematics to have at least a bit experience with some more abstract mathematics. I got this analysis book from my brother who studies physics.
     
    Last edited: Jul 28, 2016
  10. Jul 28, 2016 #9

    fresh_42

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    It will get better with time. Wait until you will finish your proofs as soon as some ##\epsilon## show up :smile:
    and all the work now, like ##\epsilon' = 2 \epsilon##, ##\epsilon' = \frac{1}{3} \epsilon## or ##\epsilon = \min\{\epsilon_1, \epsilon_2\}## will be classified under the label: "boring technical issues". Unfortunately until then, the boring stuff has to be learned, too. (see post #7)
     
  11. Jul 28, 2016 #10

    Math_QED

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    Well I do not think it's boring at all ;)
     
  12. Jul 28, 2016 #11

    PeroK

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    The other aspect of this is the powerful logical thinking required. For example, the negation of the limit. This confounds a lot of people when they first meet it. How do you say that a sequence does not have a limit of ##L##? This logical thinking is vital, even if the epsilons and deltas have been replaced by other more abstract concepts.
     
  13. Jul 28, 2016 #12

    Math_QED

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    I'm not quite sure about this one:

    Would this be:

    ∃ε > 0, ∀N: n>N => |un - L | >= ε

    or something like..

    ∀ε > 0, there is no N: n>N => |un - L | < ε

    EDIT: could someone give a look at post #8?
     
  14. Jul 28, 2016 #13

    PeroK

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    Neither, I'm afraid. Have a good think!
     
  15. Jul 28, 2016 #14

    micromass

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    Both are incorrect.
     
  16. Jul 28, 2016 #15

    Svein

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    Not only newbies. In the University I once had a very old-fashioned professor. When proving a limit, he chose an ε and a δ, and after about half an hour he ended up with ".. less than 2ε". Of course, we were satisfied, but he started the next lecture by choosing a δ half as big as he had done in the previous lecture and did the whole proof over!

    On another note, I remember a proof in multidimensional complex analysis which ended up with "... less that 10 000ε, which is small when ε is small."
     
  17. Jul 28, 2016 #16

    Math_QED

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    ∃ε > 0, ∀N: n>N ∧ |un - L | >= ε

    or: ∃ε > 0, ∀N: |un - L | >= ε => n=<N

    I used neg(a=>b) <=> a Λ neg(b) for the first and contraposition for the second. I had to look up some logic first because this is something we were not taught in high school.

    Can someone look at post #8, this is where I am stuck in the proof.
     
  18. Jul 28, 2016 #17

    PeroK

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    It's not a dumb question. The crux of the matter is that if you choose any ##\epsilon > 0##, then ##\epsilon ' = \epsilon/2 > 0##. And, for this ##\epsilon '## you can find ##N_1, N_2## such that:

    ##n > N_1 \Rightarrow |u_n - l_1| < \epsilon ' = \epsilon/2##

    And

    ##n > N_2 \Rightarrow |v_n - l_2| < \epsilon ' = \epsilon/2##

    Now, ##n > max\{N_1, N_2 \}## gives you what you want.

    Does that make sense?

    Personally, I would do this without introducing ##\epsilon '##. I'd just say that because ##\epsilon/2 > 0## we can find an ##N_1## such that:

    ##n > N_1 \Rightarrow |u_n - l_1| < \epsilon/2## etc.

    PS I can see the book did things a bit differently. I definitely don't like using ##N'## for the ##l_2## limit. Using ##N_1, N_2## seems much more logical. Especially as ##N'## has nothing to do with ##\epsilon '##
     
    Last edited: Jul 28, 2016
  19. Jul 28, 2016 #18

    Math_QED

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    In fact, the book used N1 and N2 but I wrote down the proof without the book. You are right that this is more logical. I believe your explanation helped me further. Thank you for that! I will think about it some more. In the meantime, what do you think about the negation of the limit?
     
  20. Jul 28, 2016 #19

    PeroK

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    For the negation, think about the sequence:

    ##1, 1/2, 1, 1/4, 1, 1/8, \dots##

    And why it does not converge to either ##0## or ##1## (or anything else!).
     
  21. Jul 28, 2016 #20

    Math_QED

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    Is what I wrote in post #16 wrong?
     
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